Question

A conical container, with vertex down, has a height of 6 cm and a diameter of 2 cm. It is leaking water at

Answer 1

Answer:

$\displaystyle \frac{dh}{dt}=-\frac{4}{\pi}\approx-1.2732\text{ centimeters per minute}$

The water level is dropping by approximately 1.27 centimeters per minute.

Step-by-step explanation:

Please refer to the attached diagram.

The height of the conical container is 6 cm, and its radius is 1 cm.

The container is leaking water at a rate of 1 cubic centimeter per minute.

And we want to find the rate at which the water level h is dropping when the water height is 3 cm.

Since we are relating the water leaked to the height of the water level, we will consider the volume formula for a cone, given by:

$\displaystyle V=\frac{1}{3}\pi r^2h$

Now, we can establish the relationship between the radius r and the height h. At any given point, we will have two similar triangles as shown below. Therefore, we can write:

$\displaystyle \frac{1}{6}=\frac{r}{h}$

Solving for r yields:

$\displaystyle r=\frac{1}{6}h$

So, we will substitute this into our volume formula. This yields:

$\displaystyle \begin{aligned} V&=\frac{1}{3}\pi \Big(\frac{1}{6}h\Big)^2h\\ &=\frac{1}{108}\pi h^3\end{aligned}$

Now, we will differentiate both sides with respect to time t. Hence:

$\displaystyle \frac{d}{dt}[V]=\frac{d}{dt}\Big[\frac{1}{108}\pi h^3\Big]$

The left is simply dV/dt. We can move the coefficient from the right:

$\displaystyle \frac{dV}{dt}=\frac{1}{108}\pi\frac{d}{dt}\big[h^3\big]$

Implicitly differentiate:

$\displaystyle\begin{aligned} \frac{dV}{dt}&=\frac{1}{108}\pi(3h^2\frac{dh}{dt})\\ &=\frac{1}{36}\pi h^2\frac{dh}{dt}\end{aligned}$

Since the water is leaking at a rate of 1 cubic centimeter per minute, dV/dt=-1.

We want to find the rate at which the water level h is dropping when the height of the water is 3 cm.. So, we want to find dh/dt when h=3.

So, by substitution, we acquire:

$\displaystyle -1=\frac{1}{36}\pi(3)^2\frac{dh}{dt}$

Therefore:

$\displaystyle -1=\frac{1}{4}\pi\frac{dh}{dt}$

Hence:

$\displaystyle \frac{dh}{dt}=-\frac{4}{\pi}\approx-1.2732\text{ centimeters per minute}$

The water level is dropping at a rate of approximately 1.27 centimeters per minute.